Every subset of $\bar{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\bar{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$

Case S1 for the supremum: $+\mathrm{\infty }\in A$.

Case S2 for the supremum: $+\mathrm{\infty }\mathrm{\notin }A$ and $A_{\mathbb{R}}=\mathrm{\varnothing }$, so that every element of $A$ equals $-\mathrm{\infty }$.

Case S3 for the supremum: $+\mathrm{\infty }\mathrm{\notin }A$, $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$, and $A_{\mathbb{R}}$ is bounded above in $\mathbb{R}$.

Case S4 for the supremum: $+\mathrm{\infty }\mathrm{\notin }A$, $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$, and $A_{\mathbb{R}}$ is not bounded above in $\mathbb{R}$.

Case I1 for the infimum: $-\mathrm{\infty }\in A$.

Case I2 for the infimum: $-\mathrm{\infty }\mathrm{\notin }A$ and $A_{\mathbb{R}}=\mathrm{\varnothing }$, so that every element of $A$ equals $+\mathrm{\infty }$.

Case I3 for the infimum: $-\mathrm{\infty }\mathrm{\notin }A$, $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$, and $A_{\mathbb{R}}$ is bounded below in $\mathbb{R}$.

Case I4 for the infimum: $-\mathrm{\infty }\mathrm{\notin }A$, $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$, and $A_{\mathbb{R}}$ is not bounded below in $\mathbb{R}$.

In case S1 the element $+\mathrm{\infty }$ is an upper bound of $A$, being the greatest element of $\bar{\mathbb{R}}$; and if $v$ is any upper bound of $A$ then $+\mathrm{\infty }\in A$ gives $+\mathrm{\infty }\le v$, whence $v=+\mathrm{\infty }$ by antisymmetry. So $+\mathrm{\infty }$ is the least upper bound of $A$.

In case S2 every element of $A$ equals $-\mathrm{\infty }$, so $-\mathrm{\infty }$ is an upper bound of $A$ by reflexivity; and $-\mathrm{\infty }\le v$ for every $v\in \bar{\mathbb{R}}$, being the least element. So $-\mathrm{\infty }$ is the least upper bound of $A$.

In case S3 the real number $\sigma :=\sup A_{\mathbb{R}}$ exists, and it is an upper bound of $A$ in $\bar{\mathbb{R}}$: an element of $A$ is either real, hence lies in $A_{\mathbb{R}}$ and satisfies $a\le \sigma$ in $\mathbb{R}$ and so in $\bar{\mathbb{R}}$, or equals $-\mathrm{\infty }$, which is $\le \sigma$; the value $+\mathrm{\infty }$ does not occur in $A$ in this case.

In case S4 the element $+\mathrm{\infty }$ is an upper bound of $A$; and if $v$ is an upper bound then $v\mathrm{\ne }-\mathrm{\infty }$, because fixing $a\in A_{\mathbb{R}}$, which is possible in this case, gives $a\le v$ with $a$ real and $-\mathrm{\infty }$ is below no real, while $v$ real would make $v$ a real upper bound of $A_{\mathbb{R}}$ and contradict the case hypothesis. So $v=+\mathrm{\infty }$, and $+\mathrm{\infty }$ is the least upper bound of $A$.

In case I1 the element $-\mathrm{\infty }$ is a lower bound of $A$, being least; and any lower bound $w$ satisfies $w\le -\mathrm{\infty }$ because $-\mathrm{\infty }\in A$, whence $w=-\mathrm{\infty }$ by antisymmetry. So $-\mathrm{\infty }$ is the greatest lower bound of $A$.

In case I2 every element of $A$ equals $+\mathrm{\infty }$, so $+\mathrm{\infty }$ is a lower bound of $A$ by reflexivity, and $w\le +\mathrm{\infty }$ for every $w$. So $+\mathrm{\infty }$ is the greatest lower bound of $A$.

In case I3 the real number $\iota :=\inf A_{\mathbb{R}}$ exists and is a lower bound of $A$ in $\bar{\mathbb{R}}$: an element of $A$ is either real, hence in $A_{\mathbb{R}}$ and $\ge \iota$, or equals $+\mathrm{\infty }\ge \iota$; the value $-\mathrm{\infty }$ does not occur in $A$ in this case.

In case I4 the element $-\mathrm{\infty }$ is a lower bound of $A$; any lower bound $w$ satisfies $w\mathrm{\ne }+\mathrm{\infty }$, because fixing $a\in A_{\mathbb{R}}$ gives $w\le a$ with $a$ real and $+\mathrm{\infty }$ is above no real, while $w$ real would be a real lower bound of $A_{\mathbb{R}}$ and contradict the case hypothesis. So $w=-\mathrm{\infty }$ is the greatest lower bound of $A$.

In case S3 let $v$ be any upper bound of $A$ and fix $a\in A_{\mathbb{R}}$, which is possible since $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$. From $a\le v$ with $a$ real we get $v\mathrm{\ne }-\mathrm{\infty }$, since $-\mathrm{\infty }$ is below no real. If $v=+\mathrm{\infty }$ then $\sigma \le v$ because $+\mathrm{\infty }$ is greatest. Otherwise $v$ is real, and it bounds $A_{\mathbb{R}}$ above in $\mathbb{R}$, so $\sigma \le v$ by leastness of the real supremum. Hence $\sigma$ is the least upper bound of $A$.

In case I3 let $w$ be a lower bound of $A$ and fix $a\in A_{\mathbb{R}}$. From $w\le a$ with $a$ real we get $w\mathrm{\ne }+\mathrm{\infty }$. If $w=-\mathrm{\infty }$ then $w\le \iota$; otherwise $w$ is real and bounds $A_{\mathbb{R}}$ below in $\mathbb{R}$, so $w\le \iota$. Hence $\iota$ is the greatest lower bound of $A$.

The four supremum cases are exhaustive and mutually exclusive: either $+\mathrm{\infty }\in A$, which is S1, or not, and then either $A_{\mathbb{R}}=\mathrm{\varnothing }$, which is S2, or $A_{\mathbb{R}}\mathrm{\ne }\mathrm{\varnothing }$ and it is bounded above in $\mathbb{R}$, which is S3, or it is not, which is S4. In each case a least upper bound was produced, and it is unique. The same four alternatives with $-\mathrm{\infty }$, $+\mathrm{\infty }$ and "below" in place of $+\mathrm{\infty }$, $-\mathrm{\infty }$ and "above" are I1 to I4, and in each a greatest lower bound was produced.

The agreement clause follows: a nonempty $A\subseteq \mathbb{R}$ bounded above in $\mathbb{R}$ satisfies $+\mathrm{\infty }\mathrm{\notin }A$ and $A_{\mathbb{R}}=A$, so case S3 applies and $\sup A=\sup A_{\mathbb{R}}$ is the real supremum; a nonempty $A\subseteq \mathbb{R}$ bounded below satisfies case I3 and $\inf A$ is the real infimum; and $A=\mathrm{\varnothing }$ falls under S2 and I2, giving $\sup \mathrm{\varnothing }=-\mathrm{\infty }$ and $\inf \mathrm{\varnothing }=+\mathrm{\infty }$.